Question #0e76f

1 Answer

The answer is #n=29#

Explanation:

To solve this, we can use a very known result, which is the formula to the sum of the #n# first squares. It is as follows:

#\sum_[k=1]^n k^2 = (n(n+1)(2n+1))/6#

Substituting the formula into the equation, we get

#1/(n(2n+1)) * (n(n+1)(2n+1))/6 = 5#

Simplifying:

#(n+1)/6 = 5#

All we need to do now is solve for #n#

#n+1 = 30 => n=29#//

Proof to the formula of the sum of the #n# first squares can be found here: http://mathforum.org/library/drmath/view/56920.html

Hope it helps.